Question

score: 5.25/9 answered: 6/9
question 7
write an equation for the ellipse graphed in standard form.
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Explanation:

Step1: Find the center of the ellipse

The center \((h,k)\) of the ellipse is at the midpoint of the major and minor axes. From the graph, the center is at \((3, 3)\) (since the ellipse is centered at the point where \(x = 3\) and \(y=3\) by observing the grid).

Step2: Determine the lengths of the semi - major and semi - minor axes

• For the horizontal axis (major or minor, we check the length): The ellipse extends from \(x=- 1\) to \(x = 7\) (since the leftmost point is at \(x=-1\) and the rightmost at \(x = 7\)). The length of the major (horizontal) axis is \(7-(-1)=8\), so the semi - major axis \(a=\frac{8}{2}=4\).
• For the vertical axis: The ellipse extends from \(y = 1\) to \(y=5\). The length of the minor (vertical) axis is \(5 - 1=4\), so the semi - minor axis \(b=\frac{4}{2}=2\).

Step3: Write the standard form of the ellipse equation

The standard form of an ellipse with a horizontal major axis (since \(a>b\) here, \(a = 4\), \(b = 2\)) is \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\), where \((h,k)\) is the center.

Substituting \(h = 3\), \(k = 3\), \(a = 4\), and \(b = 2\) into the equation, we get:

\(\frac{(x - 3)^{2}}{4^{2}}+\frac{(y - 3)^{2}}{2^{2}}=1\)

Simplifying the denominators: \(\frac{(x - 3)^{2}}{16}+\frac{(y - 3)^{2}}{4}=1\)

Answer:

\(\frac{(x - 3)^{2}}{16}+\frac{(y - 3)^{2}}{4}=1\)